All categories

3 years ago

The snake population in a swamp Is 1056. The population of the snakes is increasing by a factor of 28% each year.

1 answer:

6 0

a. Model the situation with an exponential function.
The exponential function that models the problem in this case is:
y = 1056 (1.28) ^ t
Where,
t: number of years

b. Find how many alligators there will be in 8 years.
For 8 years we must replace the value of t = 8 in the written equation.
We have then:
y = 1056 (1.28) ^ 8
y = 7609.28193
Nearest whole integer:
y = 7609

c. How many years will it take for the population to become 3000?

For this we should almost replace y = 3000 in the equation of part A.
We have then:
3000 = 1056 (1.28) ^ t
From here, we clear t:
(1.28) ^ t = (3000) / (1056)
We apply logarithm to both sides:
log1.28 ((1.28) ^ t) = log1.28 ((3000) / (1056))
t = log1.28 ((3000) / (1056))
t = 4.229619111 years.

You might be interested in

Solve for t and find the value for t if I= 115, P=650, r=5%

Brrunno [24]

P=650
R= 5% annually= 0.05
T=1 month=1/12 year

Sl=Prt=650*0.05*1/12= $2.70
So the answer would be 2.7 or 2.70

5 0

3 years ago

Simplify 5/13 ( 75 - 49)

Hatshy [7]

Answer:

The answer is 10.

Step-by-step explanation:

I'm certain this is correct. :D

4 0

3 years ago

How do you solve 3^5

ankoles [38]

3^5 is the same as 3*3*3*3*3. which is 243

5 0

3 years ago

Read 2 more answers

Drag each equation to the correct location on the table.

Ket [755]

Answer:

Hi! I hope this is correct please inform me if this is incorrect :)

a = 1 will have.

2a + 3 < 9 - 3a.

2a + 3 > 10 -3a.

3a - 2a + 1>2.

a = 2 will have

ANY others I did not include in a = 1 :)

Please once again tell me if this is incorrect.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$